46th Putnam 1985

Problem A2

ABC is an acute-angled triangle with area 1. A rectangle R has its vertices Ri on the sides of the triangle, R1 and R2 on BC, R3 on AC and R4 on AB. Another rectangle S has it vertices on the sides of the triangle AR3R4, two on R3R4 and one one each of the other two sides. What is the maximum total area of R and S over all possible choices of triangle and rectangles?



Answer: 2/3.

Let k = AR3/AC. The triangle AR4R3 is similar to ABC with all dimensions a factor k smaller. So its area is k2. Similarly, the two triangles BR4R1 and CR3R2 together have area (1-k)2. So the area of the bottom rectangle is 1 - k2 - (1-k)2 = 2k(1-k). Defining h in a similar way, the area of the top rectangle is 2h(1-h) times the area of triangle AR4R3 = 2h(1-h)k2.

For any given choice of k, we maximise the area of the top rectangle by taking h = 1/2. The area of the two rectangles together is then 2k(1-k) + k2/2 = 2/3 - 3/2(k - 2/3)2. This evidently has maximum value 2/3, obtained by taking k = 2/3. Note that the maximum is independent of the choice of triangle.



46th Putnam 1985

© John Scholes
7 Jan 2001